In: Chemistry
In-Class Problem/Take-Home Problem 6: For the following aqueous rxn, 0.15M C and 0.087M D are
allowed to react. At equilibrium, only 0.009M D remains. Calculate K and all of the equilibrium
concentrations.
2A + B C + 3D
2 A + B <-----------------------------> C + 3 D
0 0 0.15 0.087 ----------------------> initial
+2x x -x -3x --------------------------> change
2x x 0.15-x 0.087-3x ----------------------> equilibrium
but it is given D at equilibrium 0.087-3x = 0.009
x = 0.026
equilibrium concentrations :
[A] = 2x = 0.052 M
[B] = x = 0.026 M
[C] = 0.15-x = 0.12 M
[D] = 0.009 M
equilibrium constant K = products / reactnats
= (0.12) (0.009)^3 / (0.052)^2 (0.026)
= 1.29 x 10^-3